#### Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 15

$y^{2}=4 a(x+a)$  is a solution of differential equation

Hint:

Differentiate the solution and modify the given differential equation

Given:

$y^{2}=4 a(x+a)$

Solution:

Differentiating on both sides with respect to $x$

\begin{aligned} &\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(4 a(x+a)) \\\\ &2 y \frac{d y}{d x}=4 a \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{4 a}{2 y} \\\\ &\frac{d y}{d x}=\frac{2 a}{y} \end{aligned}                                ...............(i)

Put value of equation (i) in given problem as follows

\begin{aligned} &y\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=2 x \frac{d y}{d x} \\\\ &L H S=y\left[1-\left(\frac{d y}{d x}\right)^{2}\right] \end{aligned}

\begin{aligned} &=y\left[1-\left(\frac{2 a}{y}\right)^{2}\right] \\\\ &=y-\frac{4 a^{2} y}{y^{2}} \\\\ &=y-\frac{4 a^{2}}{y} \end{aligned}                        ..............(ii)

\begin{aligned} &R H S=2 x \frac{d y}{d x} \\\\ &=2 x\left(\frac{2 a}{y}\right) \\\\ &=\frac{4 a x}{y} \end{aligned}                                     ..............(iii)

Now,  $y^{2}=4 a(x+a)$

\begin{aligned} &y^{2}=4 a x+4 a^{2} \\\\ &\frac{y^{2}-4 a^{2}}{4 a}=x \end{aligned}                                    ...........(iv)

Put value of equation (iv) in equation (iii)

\begin{aligned} &=\frac{4 a}{y}\left(\frac{y^{2}-4 a^{2}}{4 a}\right) \\\\ &=\frac{y^{2}}{y}-\frac{4 a^{2}}{y} \\\\ &=y-\frac{4 a^{2}}{y} \end{aligned}                                ............(v)

From (ii) and (v)

$L H S=R H S$

Thus, $y^{2}=4 a(x+a)$ is a solution of differential equation.